Integration by Substitution Solutions
Note:
in (1) through (6), we find the answer by setting
du
equal to the
derivative of the function, multiplied by
dx
.
1.
u
= 4
x
+ 1
du
= 4
dx
2.
u
=
x
4
+ 2
du
= 4
x
3
dx
3.
u
= 3
x
2
+ 1
du
= 6
xdx
4.
u
= ln(
x
)
du
=
1
x
dx
5.
u
= 1
/x
=
x
-
1
du
=
-
x
-
2
dx
6.
u
=
e
x
+ 2
du
=
e
x
dx
7.
u
= 4
x
+ 1
du
= 4
dx
du/
4 =
dx
R
(4
x
+ 1)
4
dx
=
R
u
4
du
4
=
1
4
R
u
4
du
=
1
4
·
u
5
5
+
C
=
1
20
(4
x
+ 1)
5
+
C
8.
u
=
-
2
x
+ 3
du
=
-
2
dx
-
du/
2 =
dx
R
e
-
2
x
+3
dx
=
R
e
u
· -
du
2
=
-
1
2
R
e
u
du
=
-
1
2
e
u
+
C
=
-
1
2
e
-
2
x
+3
+
C
9.
Note:
in problems like this one, it may not be immediately obvious
what to set
u
equal to (e.g., here, you might set
u
= 5
x
3
or
u
=
x
4
+2). You
can try trial and error, by setting
u
equal to something and trying to solve
1
2
the problem, and then starting over if you can’t cancel out all of the ‘
x
’s;
another tip is to look for a function whose derivative you see in the integral.
So if you set
u
=
x
4
+ 2, the derivative is 4
x
3
, which is similar to the 5
x
3
we already see in the integral; this is a good indicator that
u
=
x
4
+ 2 is the
right choice.
